/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 A specimen of oil having an init... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A specimen of oil having an initial volume of 600 \(\mathrm{cm}^{3}\) is subjected to a pressure increase of \(3.6 \times 10^{6} \mathrm{Pa}\) , and the volume is found to decrease by 0.45 \(\mathrm{cm}^{3} .\) What is the bulk modulus of the material? The compressibility?

Short Answer

Expert verified
Bulk modulus is \(4.8 \times 10^9 \ \text{Pa}\); compressibility is \(2.08 \times 10^{-10} \ \text{Pa}^{-1}\).

Step by step solution

01

Understand the Bulk Modulus Formula

The bulk modulus, denoted as \( B \), is defined as \( B = -\frac{\Delta P}{\frac{\Delta V}{V_0}} \), where \( \Delta P \) is the change in pressure, \( \Delta V \) is the change in volume, and \( V_0 \) is the initial volume.
02

Substitute the Known Values

Substitute the given values into the formula: \( \Delta P = 3.6 \times 10^6 \ \text{Pa} \), \( \Delta V = -0.45 \ \text{cm}^3 \) (negative because the volume decreases), and \( V_0 = 600 \ \text{cm}^3 \). This gives the equation \( B = -\frac{3.6 \times 10^6}{\frac{-0.45}{600}} \).
03

Simplify the Equation

Simplify the denominator \( \frac{-0.45}{600} = -0.00075 \), thus the bulk modulus \( B = \frac{3.6 \times 10^6}{0.00075} \).
04

Calculate the Bulk Modulus

Calculate the bulk modulus \( B = 4.8 \times 10^9 \ \text{Pa} \). This indicates the resistance of the oil to volume change under pressure.
05

Understand Compressibility

Compressibility, denoted as \( \beta \), is the reciprocal of the bulk modulus: \( \beta = \frac{1}{B} \). This value represents how easily the material can be compressed.
06

Calculate Compressibility

Calculate the compressibility \( \beta = \frac{1}{4.8 \times 10^9} \). Simplify to find \( \beta \approx 2.08 \times 10^{-10} \ \text{Pa}^{-1} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Compressibility
Compressibility is a fascinating concept that tells us how much a material can be compressed. Think of it as the opposite of rigidity. Unlike solids, liquids and gases aren't as stiff, which means they can be squeezed or pressed into smaller spaces more easily. This property of a material is measured by something called compressibility, denoted as \( \beta \).
The science behind this is quite straightforward. Compressibility is defined as the reciprocal of the bulk modulus \( B \). The formula is:
  • \( \beta = \frac{1}{B} \)
Basically, if the bulk modulus is high, the compressibility is low. This means the material doesn't like to be compressed. Conversely, a low bulk modulus means higher compressibility, indicating that the substance can be easily pressed into a smaller volume.
In our exercise, we calculated the compressibility of the oil to be approximately \( 2.08 \times 10^{-10} \ \text{Pa}^{-1} \). This tells us that oil, like many liquids, has low compressibility compared to gases, but definitely more than rigid solids like metals.
Pressure Change
Understanding pressure change is key in physics, especially when dealing with fluids. Pressure is defined as the force applied perpendicular to the surface of an object over a specific area. When this pressure changes, it can lead to changes in volume for a compressible material.
For our oil example, the pressure increase \( \Delta P \) is given as \( 3.6 \times 10^6 \ \text{Pa} \). When pressure is applied, the molecules in the oil move closer together, causing the volume to decrease. This change in pressure is critical for calculating compressibility and bulk modulus.
  • If \( \Delta P \) is positive, the pressure has increased.
  • If \( \Delta P \) is negative, the pressure has decreased.
Knowing how pressure affects a material helps engineers and scientists design equipment and processes that can withstand or utilize these changes.
Volume Change
Volume change is an important factor to understand when studying how materials react to different conditions. Volume is how much space an object or substance occupies, and when external conditions change, so can the volume.
In the oil problem, we noticed a volume decrease of \(-0.45 \ \text{cm}^3 \). The negative sign indicates that the volume is reduced rather than expanded. This happens when pressure increases and the molecules in the material get pushed closer together.
  • Positive volume change means expansion.
  • Negative volume change indicates contraction.
Understanding these changes is essential because it helps predict how materials will behave under different circumstances, like high pressure or temperature, which is crucial in many fields, including aerospace, automotive design, and even cooking!
Physics Problem-Solving
Physics problem-solving requires a systematic approach to unravel complex scenarios into comprehensible solutions. Here's a simplified way to tackle such problems:
  • **Understand the Problem:** Break it down to its core components. Identify given values and quantities to be calculated.
  • **Select the Right Formulas:** Use formulas relevant to the topic, like the bulk modulus \( B \) for volume changes under pressure.
  • **Substitute and Simplify:** Plug the given values into the formulas and simplify. This involves arithmetic and algebraic manipulation.
  • **Interpret:** Understand what the answer means in terms of the real-world scenario you're analyzing.
In our case, applying these steps helped us find both the bulk modulus and compressibility. Practicing this structured method improves skills in solving various physics challenges.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

BIO Biceps Muscle. A relaxed biceps muscle requires a force of 25.0 \(\mathrm{N}\) for an elongation of 3.0 \(\mathrm{cm}\) ; the same muscle under maximum tension requires a force of 500 \(\mathrm{N}\) for the same elongation. Find Young's modulus for the muscle tissue under each of these conditions if the muscle is assumed to be a uniform cylinder with length 0.200 \(\mathrm{m}\) and cross-sectional area 50.0 \(\mathrm{cm}^{2} .\)

A 350 -N, uniform, 1.50 -m bar is suspended horizontally by two vertical cables at each end. Cable \(A\) can support a maximum tension of 500.0 \(\mathrm{N}\) without breaking, and cable \(B\) can support up to 400.0 \(\mathrm{N} .\) You want to place a small weight on this bar. (a) What is the heaviest weight you can put on without breaking either cable, and (b) where should you put this weight?

An engineer is designing a conveyor system for loading hay bales into a wagon (Fig. P11.81). Each bale is 0.25 \(\mathrm{m}\) wide, 0.50 \(\mathrm{m}\) high, and 0.80 \(\mathrm{m}\) long (the dimension perpendicular to the plane of the figure), with mass 30.0 kg. The center of gravity of each bale is at its geometrical center. The coefficient of static friction between a bale and the conveyor belt is 0.60 , and the belt moves with constant speed. (a) The angle \(\beta\) of the conveyor is slowly increased. At some critical angle a bale will tip (if it doesn't slip first), and at some different critical angle it will slip (if it doesn't tip first). Find the two critical angles and determine which happens at the smaller angle. (b) Would the outcome of part (a) be different if the coefficient of friction were 0.40\(?\)

Suppose that you can lift no more than 650 \(\mathrm{N}\) (around 150 lb) unaided. (a) How much can you lift using a \(1.4-\mathrm{m}\) -long wheelbarrow that weighs 80.0 \(\mathrm{N}\) and whose center of gravity is 0.50 \(\mathrm{m}\) from the center of the wheel (Fig. E11.16)? The center of gravity of the load carried in the wheelbarrow is also 0.50 \(\mathrm{m}\) from the center of the wheel. (b) Where does the force come from to enable you to lift more than 650 \(\mathrm{N}\) using the wheelbarrow?

A uniform strut of mass \(m\) makes an angle \(\theta\) with the horizontal. It is supported by a frictionless pivot located at one- third its length from its lower left end and a horizontal rope at its upper right end. A cable and package of total weight \(w\) hang from its upper right end. (a) Find the vertical and horizontal components \(V\) and \(H\) of the pivot's force on the strut as well as the tension \(T\) in the rope. (b) If the maximum safe tension in the rope is 700 \(\mathrm{N}\) and the mass of the strut is \(30.0 \mathrm{kg},\) find the maximum safe weight of the cable and package when the strut makes an angle of \(55.0^{\circ}\) with the horizontal. (c) For what angle \(\theta\) can no weight be safely suspended from the right end of the strut?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.